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Lesson 9 - R
Chapter 9 Review
Objectives
• Summarize the chapter
• Define the vocabulary used
• Complete all objectives
• Successfully answer any of the review exercises
Vocabulary
• None new
Determine the Appropriate Confidence Interval to Construct
Which parameter
are we
estimating
Standard Deviation, σ,
or variance σ²
1) normal population
2) no outliers
yes
Mean, μ
Proportion p
Assumptions Met?
1) n ≤ 0.05N
2) np(1-p) ≥ 10
yes
σ known?
yes
no
Compute χ²-interval
yes
Compute Z-interval
n≥30
yes & σ
no
no
1) apx normal population
2) no outliers
n≥30
yes
yes & s
no
Compute Z-interval
Nonparametics
Compute t-interval
Chapter 9 – Section 1
If the sample mean is 9, which of these
could reasonably be a confidence
interval for the population mean?
1)
2)
3)
4)
92
(3, 6)
(7, 11)
(0, ∞)
Chapter 9 – Section 1
If the population standard deviation σ = 5
and the sample size n = 25, then the
margin of error for a 95% normal
confidence interval is
1)
2)
3)
4)
1
2
5
25
Chapter 9 – Section 2
A researcher collected 15 data points that
seem to be reasonably bell shaped. Which
distribution should the researcher use to
calculate confidence intervals?
1)
2)
3)
4)
A t-distribution with 14 degrees of freedom
A t-distribution with 15 degrees of freedom
A general normal distribution
A nonparametric method
Chapter 9 – Section 2
What issue do we have in calculating
σ / √ n when the population standard
deviation is not known?
1) There are no issues
2) We do not know which value to use for n
3) We do not know how to calculate the
sample mean
4) We do not know which value to use for σ
Chapter 9 – Section 3
A study is trying to determine what
percentage of students drive SUVs. The
population parameter to be estimated is
1)
2)
3)
4)
The sample mean
The population proportion
The standard error of the sample mean
The sample size required
Chapter 9 – Section 3
A study of 100 students to determine a
population proportion resulted in a
margin of error of 6%. If a margin of error
of 2% was desired, then the study should
have included
1)
2)
3)
4)
200 students
400 students
600 students
900 students
Chapter 9 – Section 4
Which probability distribution is used to
compute a confidence interval for the
variance?
1)
2)
3)
4)
The normal distribution
The t-distribution
The α distribution
The chi-square distribution
Chapter 9 – Section 4
If the 90% confidence interval for the
variance is (16, 36), then the 90%
confidence interval for the standard
deviation is
1)
2)
3)
4)
(4, 6)
(8, 18)
(160, 360)
Cannot be determined from the information
given
Chapter 9 – Section 5
Which of the following methods are used
to estimate the population mean?
1) The margin of error using the normal
distribution
2) The margin of error using the t-distribution
3) Nonparametric methods
4) All of the above
Chapter 9 – Section 5
A professor wishes to compute a
confidence interval for the average
percentage grade in the class. Which
population parameter is being studied?
1)
2)
3)
4)
The population mean
The population proportion
The population variance
The population standard deviation
Summary and Homework
• Summary
– We can use a sample {mean, proportion, variance,
standard deviation} to estimate the population
{mean, proportion, variance, standard deviation}
– In each case, we can use the appropriate model to
construct a confidence interval around our
estimate
– The confidence interval expresses the confidence
we have that our calculated interval contains the
true parameter
• Homework:
pg 497 – 501; 1, 3, 8, 9, 15, 23, 24
Homework
1: (α/2=0.005, df=18-1=17) read from table: 2.898
3: (α/2=0.025, df=22-1=21) read from table: 10.283, 35.479
8: a) n>30 large sample (σ known)
b) (α/2=0.03) Z=1.88 [315.15, 334.85]
c) (α/2=0.01) Z=2.326 [312.81, 337.19]
d) (α/2=0.025) Z=1.96 n > 147.67
9: a) large sample size allows for x-bar to be normally distributed
from a non-normal (skewed data distribution: mean vs median)
b) (α/2=0.05) t=1.646 MOE=0.298 [12.702, 13.298]
15: a) x-bar = 3.244, s = 0.487
b) yes
c) (α/2=0.025) [2.935, 3.553]
d) (α/2=0.005) [2.807, 3.681]
e) (α/2=0.005) [0.312, 1.001]
23: same because t-dist is symmetric
24: t-dist, because the tails are larger